1. **Problem:** Evaluate the integral $$\int \frac{3}{4 + t^2} \, dt$$. 2. **Formula:** Recall the integral formula for $$\int \frac{1}{a^2 + x^2} \, dx = \frac{1}{a} \arctan\left(\frac{x}{a}\right) + C$$ where $$a > 0$$. 3. **Apply the formula:** Here, $$a^2 = 4$$ so $$a = 2$$. 4. **Rewrite the integral:** $$\int \frac{3}{4 + t^2} \, dt = 3 \int \frac{1}{4 + t^2} \, dt = 3 \int \frac{1}{2^2 + t^2} \, dt$$. 5. **Evaluate:** Using the formula, $$\int \frac{1}{2^2 + t^2} \, dt = \frac{1}{2} \arctan\left(\frac{t}{2}\right) + C$$. 6. **Multiply by 3:** $$3 \times \frac{1}{2} \arctan\left(\frac{t}{2}\right) + C = \frac{3}{2} \arctan\left(\frac{t}{2}\right) + C$$. **Final answer:** $$\boxed{\int \frac{3}{4 + t^2} \, dt = \frac{3}{2} \arctan\left(\frac{t}{2}\right) + C}$$